Problem 29 - Intermediate Algebra

Question

Solve each equation. $$ \frac{6}{x}+\frac{40}{x+5}=7 $$

Step-by-Step Solution

Answer

The solutions are \( x = 3 \) and \( x = -\frac{10}{7} \).

Step 1: Identify the Least Common Denominator (LCD)

The given equation is \( \frac{6}{x} + \frac{40}{x+5} = 7 \). To combine the fractions on the left side, we need a common denominator. The LCD of \( x \) and \( x+5 \) is \( x(x+5) \).

Step 2: Multiply Through by the LCD

Multiply every term in the equation by the LCD \( x(x+5) \) to eliminate the fractions: \[ x(x+5) \left( \frac{6}{x} \right) + x(x+5) \left( \frac{40}{x+5} \right) = 7 \, x(x+5) \]This simplifies to:\[ 6(x+5) + 40x = 7x(x+5) \]

Step 3: Expand and Simplify

Distribute on both sides:\[ 6x + 30 + 40x = 7x^2 + 35x \]Combine like terms on the left side:\[ 46x + 30 = 7x^2 + 35x \]

Step 4: Move All Terms to One Side

Rearrange and move all terms to one side of the equation to form a quadratic equation:\[ 0 = 7x^2 + 35x - 46x - 30 \]Combine like terms:\[ 0 = 7x^2 - 11x - 30 \]

Step 5: Solve the Quadratic Equation

Factor the quadratic equation if possible or use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 7 \), \( b = -11 \), and \( c = -30 \) to find the solutions:\[ x = \frac{-(-11) \pm \sqrt{(-11)^2 - 4\cdot7\cdot(-30)}}{2\cdot7} \]\[ x = \frac{11 \pm \sqrt{121 + 840}}{14} \]\[ x = \frac{11 \pm \sqrt{961}}{14} \]\[ x = \frac{11 \pm 31}{14} \]

Step 6: Calculate the Roots

Calculate the values of \( x \):\[ x = \frac{11 + 31}{14} = \frac{42}{14} = 3 \]\[ x = \frac{11 - 31}{14} = \frac{-20}{14} = -\frac{10}{7} \]

Step 7: Check for Extraneous Solutions

Ensure that neither solution causes a division by zero in the original equation. Both solutions \( x = 3 \) and \( x = -\frac{10}{7} \) are valid since neither results in a zero in the denominators.

Key Concepts

Quadratic Equations

Quadratic equations are fundamental in algebra. These equations take the form \( ax^2 + bx + c = 0 \) where \( a \), \( b \), and \( c \) are constants, and \( a \) is not zero. In simple terms, they describe a parabola—a U-shaped curve. To solve quadratic equations, you can utilize:

• Factoring - Splitting the equation into simpler expressions that, when multiplied, give the original equation.
• The Quadratic Formula - An essential tool given by \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), which gives the roots of the equation even if they aren't integers.
• Completing the Square - Rewriting the equation as a perfect square trinomial to find the roots.

In solving the featured exercise, we used the quadratic formula due to the nature of our quadratic equation, \( 7x^2 - 11x - 30 = 0 \). With \( a = 7 \), \( b = -11 \), and \( c = -30 \), we substituted these into the formula to find our solutions.

Least Common Denominator

When adding or subtracting fractions, finding the least common denominator (LCD) is crucial. It allows you to combine fractions into a single expression. The LCD is the smallest number that both denominators can divide without leaving a remainder. In our problem, the equation involves the fractions \( \frac{6}{x} \) and \( \frac{40}{x+5} \).
The denominators here are \( x \) and \( x+5 \). The LCD in such cases is simply the product \( x(x+5) \). Multiplying each term by the LCD will eliminate the fractions, making it easier to work with them algebraically.
By performing this multiplication, we cleared the fractions from the equation, setting the stage for solving it as a simpler quadratic equation.

Fractional Equations

Fractional equations involve one or more fractions and require careful handling to avoid mistakes. The key is to eliminate the fractions by multiplying through by their LCD, which was illustrated in our solution. This approach transforms a fractional equation into one that is likely easier to manipulate, usually simplifying to a linear or quadratic equation.
The original exercise, \( \frac{6}{x} + \frac{40}{x+5} = 7 \), needed this treatment. Once the fractions were cleared, the equation became more straightforward, and we could solve it like a regular algebraic equation.
It's important to verify any solutions derived, ensuring no denominators become zero in the process—this checks for extraneous solutions, ensuring the solutions are valid for the original equation.